Talk:xkcd: volume 0
How the pages are counted[edit]
The pages are counted in something akin to a trinary number system: the only digits used are 0, 1, and 2. But 2 always rolls over, even if there are digits behind it. So:
| Number (xkcd v0) | Number in denary |
| 1 | 1 |
| 2 | 2 |
| 10 | 3 |
| 11 | 4 |
| 12 | 5 |
| 20 | 6 |
| 100 | 7 |
| 101 | 8 |
| 102 | 9 |
| 110 | 10 |
| 111 | 11 |
| 112 | 12 |
| 120 | 13 |
| 200 | 14 |
| 1000 | 15 |
How would someone do a base conversion between "xkcd trinary" and denary? I was thinking to do something akin to (value in "normal trinary")-(value of the previous digit in the same trinary) for each magnitude in xkcd trinary, but I can't figure out how rigorous it would be, much less accurate. Any ideas? 108.162.241.141 23:33, 19 June 2024 (UTC)
Well, the recent edits up to this point don't explain much at all. XORing some numbers, but not others? What about the non-numbers (non-hexadecimal characters, "M", or even words, "Dumper")? Can I request a proper rewrite of those facts to actually make some sort of sense without apparently being selective or just plain consisting of Insane Troll Logic?
The issue with finding info is that the XKCD forum depicting the solving process doesn't really exist. I've kinda had to re-solve entire parts just for the explanation. 82.132.237.136 13:40, 23 December 2025 (UTC)
